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Examine the function f(x,y) = y2 + 4xy + 3x2 + x3 for extreme values.
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We have, $f(x,y)=y^2 + 4xy + 3x^2 + x^3$

Step I:

$f_x = 4y+6x+3x^2$

$f_y = 2y+4x$

$f_{xx} = 6 + 6x$

$f_{xy} = 4$

$f_{yy}=2$

 

Step II:

We now solve $f_x = 0, \ f_y=0$ simultaneously,

$4y+6x+3x^2 = 0 $ and $2y+4x=0$

 

Putting $2y=-4x$ in first equation

$3x^2 - 2x = 0 \ \therefore x(3x-2) = 0$

$\therefore x=0$$x=2/3$ or   When $x=0, \ y=0$ and when $x=\dfrac {2}{3}, \ y=-\dfrac {4}{3}$ $\therefore (0,0), \ \left ( \dfrac {2}{3}, - \dfrac {4}{3} \right )$ are stationary points. **Step III**: (i) When $x=0, \ y=0$ $r=f_{xx}=6+6x=6+6(0)=6$ $s=f_{xy}=4$ $t=f_{yy}=2$ \(\therefore rt-s^2 = 12 - 16 \lt 0\) We reject this pair (ii) When $x=\dfrac {2}{3}, \ y=-\dfrac {4}{3}$ $r=f_{xx}=10$ $s=f_{xy}=4$ $t=f_{yy}=2$ \(\therefore rt-s^2 = 20 -16 \gt 0\) $f(x,y)$ is stationary at $x=\dfrac {2}{3}, \ y=-\dfrac {4}{3}$ But \(r=10\gt0\) Hence, $f(x,y)$ is minimum at $x=\dfrac {2}{3}, \ y=-\dfrac {4}{3}$

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